$\begin{align} & \text{Solve the following equation by using Remainder theorem} \\ & \text{ }{{x}^{3}}-15{{x}^{2}}+74x-120=0 \\\end{align}$
$\begin{align} & \text{Let }f(x)={{x}^{3}}-15{{x}^{2}}+74x-120 \\ & \text{Using trial and error, let }x=4 \\ & f(4)={{4}^{3}}-15{{(4)}^{2}}+74(4)-120 \\ & f(4)=64-240+296-120=0 \\ & f(4)=0 \\ & \therefore x-4\text{ is a factor of }f(x) \\ & \text{Using long division } \\ & x-4\overset{{{x}^{2}}-11x+30}{\overline{\left){\begin{align} & {{x}^{3}}-15{{x}^{2}}+74x-120 \\ & \underline{{{x}^{3}}-4{{x}^{2}}\text{ }} \\ & \text{ }-11{{x}^{2}}+74x-120 \\ & \text{ }\underline{-11{{x}^{2}}-44x\text{ }} \\ & \text{ }30x-120 \\ & \text{ }\underline{\text{ }30x-120} \\ & \text{ }-------- \\\end{align}}\right.}} \\ & f(x)=({{x}^{2}}-11x+30)(x-4) \\ & f(x)=(x-6)(x-5)(x-4) \\ & \therefore \text{When }f(x)=0 \\ & (x-6)(x-5)(x-4)=0 \\ & x=6\text{ or }x=5\text{ or }x=4 \\\end{align}$